Limit Interchange for double sequence

Question

I made the following claim and gave it a proof during my winter break.

---

Given four hypothesis

1. $X=\mathbb{N},Y$ is a metric space.
2. $\{f_n\}$ uniformly converge to $f$.
3. $\lim_{k\to \infty}f(k)$ exists.
4. $\lim_{n\to \infty}\lim_{k\to \infty}f_n(k)$ exists.

We have \begin{align*} \lim_{k\to \infty}f(k)=\lim_{n\to \infty}\lim_{k\to \infty}f_n(k) \end{align*}

---

Proof:

We wish to prove that \begin{align*} \forall \epsilon , d(\lim_{k\to \infty}f(k),\lim_{n\to \infty}\lim_{k\to \infty}f_n(k))<\epsilon \end{align*} Fix $\epsilon$. Because $\lim_{k\to \infty}f_n(k) \to \lim_{n\to \infty}\lim_{k\to \infty}f_n(k)$ as $n\to \infty$, we know there exists $N$ such that \begin{align*} \forall n>N, d(\lim_{k\to \infty}f_n(k),\lim_{n\to \infty}\lim_{k\to \infty}f_n(k))<\frac{\epsilon}{4} \end{align*} Also, because $\{f_n\}$ uniformly converge to $f$, we know there exists $N'$ such that \begin{align*} \forall n>N', \forall k \in X, d(f_n(k),f(k))<\frac{\epsilon}{4} \end{align*} Let $m$ be a natural number greater than $\max \{N,N'\}$.\

Now, because $f(k)\to \lim_{k\to \infty}f(k)$ as $k\to \infty$, we know there exists $K$ such that \begin{align*} \forall k>K, d(f(k),\lim_{k\to \infty}f(k))<\frac{\epsilon}{4} \end{align*} Also, because $f_m(k)\to \lim_{k\to \infty}f_m(k)$ as $k\to \infty$, we know there exists $K'$ such that \begin{align*} \forall k>K', d(f_m(k),\lim_{k\to \infty}f_m(k))<\frac{\epsilon}{4} \end{align*} Let $j>\max \{K,K'\}$. We now have \begin{align*} d(\lim_{k\to \infty}f(k),\lim_{n\to \infty}\lim_{k\to \infty}f_n(k))&\leq d\big(\lim_{k\to \infty}f(k),f(j)\big)+d\big(f(j),f_m(j)\big)+d\big(f_m(j),\lim_{k\to \infty}f_m(k)\big)\\ &+d\big(\lim_{k\to \infty}f_m(k),\lim_{n\to \infty}\lim_{k\to \infty}f_n(k)\big)<\epsilon \end{align*}

---

I have two questions in mind, and I appreciate help in all forms.

1. Is the proof correct?

2. Does the theorem still hold true if we drop the third hypothesis? That is, from the below three hypotheses.

   1. $X=\mathbb{N},Y$ is a metric space.
   2. $\{f_n\}$ uniformly converge to $f$.
   3. $\lim_{n\to \infty}\lim_{k\to \infty}f_n(k)$ exists.

Can we deduce that $\lim_{k\to\infty} f(k)$ exists?

---

The motivation for me to ask this problem is that I know there are examples where $\lim_{k\to\infty}f(k)$ exists, but $\lim_{k\to \infty}f_n(k)$ do not exist for all $n\in\mathbb{N}$, and I for now can't come up with any example where $\lim_{n\to\infty}\lim_{k\to\infty} f_n(k)$ exists but $\lim_{k\to\infty} f(k)$ does not exists. Again, any help would be appreciated, and please notice that $Y$ need not be complete.

Answer 1

Your proof seems correct to me, and here is a proof of your conjecture that we can drop the 3rd hypothesis:

Assuming $\lim_{k\to\infty}f_n(k)= \ell_n$, $\lim_{n\to\infty}\ell_n= \ell$, and $f_n\to f$ uniformly, we shall prove that $\lim_{k\to\infty}f(k)=\ell$.

Let $\epsilon>0$. For $m$ large enough, we have both $d(\ell_m,\ell)<\epsilon/3$ and $\forall k, d(f_m(k),f(k))<\epsilon/3$. Fix such an $m$. There exists now $K$ such that $\forall k\ge K, d(f_m(k),\ell_m)<\epsilon/3$ and therefore, $d(f(k),\ell)<\epsilon$.